2. Water FilmA plate of glass (n = 1.5) is placed over a flat plate of plastic (n = 1.2). A thin film of water (n = 1.33) is trapped between the plates as illustrated in the diagram below.Glass (n = 1.5)Plastic (n = 1.2)Side view2 Water film(n = 1.33)Light of frequency f = 4.8 x 10¹4 Hz is incident from above at essentially normal incidence. The locations of the dark and bright fringes, as viewed from above, are indicated in thetop-view diagram below.Top viewAIn this problem, ignore reflections from the top surface of the glass plate and the bottom surface of the plastic.a. Are the plates in contact at point A (where the plate appears dark) or at point B (where the plate appears bright)? Explain your reasoning.b. Determine the thickness of the gap between the plates at the end where the plates are not in contact. (Express your answer in mm.) Show all work.c. Suppose that the film of water between the plates were replaced with a film of air.i. Would a bright or dark fringe be seen at the point where the two plates are in contact? Explain your reasoning.ii. When the film of water between the plates is replaced by a film of air, does the distance between adjacent bright fringes increase, decrease, or stay the same? Explain yourreasoning.

(a) The refractive index of water is less than refractive index of glass. Therefore, there is phase difference from this surface. The refractive index of water is more than refractive index of plastic. Therefore, there is no phase difference in light that is reflected from top surface of plastic. So interference occurs from reflected wave from bottom of glass plate and top of plastic plate. There is no phase difference, ∆ϕ=0. Therefore, it is consructive interference. The glass and plastic plate is not in contact, so there are dark fringes having minimum frequency. (b) For destructive interference 2nt=m+12λ where t is thickness, n is refractive index For minima, m is equal o zero. 2nt=0+12λ't=λ4 n The speed of light is c=fλ where f is frequency The thickness becomes t=c4 nft=3×108 m/s4×1.33×4.8×1014 Hzt=1.175×10-7 m (c) (i) When the film of water is replaced by film of air, the light reflects from top of plastic layer and has a phase difference of pie, therefore it is destructive interference. So, it will be a dark fringe at contact point. (ii) For adjacent bright fringes, 2nt=mλ For air, refractive index is 1. t=mλ2 For, water refractive index is 1.33 t=mλ2×1.33t=mλ2.66 Therefore, thickness for air is greater than thickness of water. So, air bright fringes are far apart than before.

Answered: A plate of glass (n = 1.5) is placed…